Develop Reference

TCP socket connection over internet

I am doing a project for which connection between server and client is required.
I did it by adding TCP sockets.
Here is the code fraction :
Server:
ServerSocket welcomeSocket = new ServerSocket(80);
while(true)
{
Socket connectionSocket = welcomeSocket.accept();
WorkerThread wt = new WorkerThread(connectionSocket, id);
Thread t = new Thread(wt);
t.start();
workerThreadCount++;
}
Client :
Socket skt = new Socket("192.168.0.108", 80); // The IP address is from cmd->ipconfig/all-> IPv4 Address
outToServer = new PrintWriter(skt.getOutputStream(), true);
inFromServer = new BufferedReader(new InputStreamReader(skt.getInputStream()));
It all works when both ends are in same device/under same WiFi.But I don't understand what to do for creating connection over internet.
Please help with clear steps.

Here:
Socket skt = new Socket("192.168.0.108", 80);
That is local address. If you want to have a server that is reachable on the internet, then that server needs to have its global public IP address!
In other words: you have to make sure that the server can be reached from the internet somehow. For example by turning to some service provider that hosts servers that you can then equip with your code!
The whole purpose of 192.168 addresses is to be defined only in a local subnet.
Alternatively, you have to check if your ISP has a service where the ISP assigns an IP address to your connection, and that allows calls from the internet to go to your "place".
Meaning: when you want to receive phone calls, you need a phone that is connected to the phone net!

In order to connect to a socket over WAN, you must port forward that port to your local device. This can be done in your routers' settings.
192.168.0.108 --> That's your local IP-address.
This can be used on your local network without any requirements for port forwarding whatsoever. However, to use it over WAN, execute the following steps:
Step 1: Search for your routers' model number and port forwarding on Google on how-to forward port 80 to your local IP-address. Warning: use a static IP-address on your local device to prevent your IP from changing after a reboot.
Step 2: Go to a website like IP Chicken and find your external IP-address.
You can then connect to your socket using:
Socket skt = new Socket("[EXTERNALIP]", 80);
Please be noticed: unless you have a business network, your external IP-address will probably change from time to time.

How to bind java.net.MulticastSocket to localhost

I would like to bind a MulticastSocket to the address 127.0.0.1 (Socket should only be reachable within the current host) but with the following code example i got a
java.net.SocketException: Network is unreachable: Datagram send failed exception
Is there a way to fix the problem? Here is my code
int port = 6677;
InetAddress group = InetAddress.getByName("232.0.1.10");
try(MulticastSocket s = new MulticastSocket(new InetSocketAddress(InetAddress.getByName("127.0.0.1"),port))){
String msg = "Hello";
s.joinGroup(group);
DatagramPacket hi = new DatagramPacket(msg.getBytes(), msg.length(),group, port);
s.send(hi);
}

Multicast is a little odd when compared to traditional UDP communication. The whole point is to share data on a known "channel", simultaneously, to anyone who wants access. This sharing is "signaled" to the network by using an IP address in the range 224.0.0.1 to 239.255.255.255. If you try to bind to 127.0.0.1, you just aren't doing Multicast anymore. And if you take a minute and think about it, that makes sense - you can't share the internal interface with other computers.

Chat using TCP in Java

I am making a chat in Java which uses a TCP protocol.
I have a client and a server side.
To send a message to another user, I have to send the message to the server through my client, and the server has to send it to another client.
The server holds the addresses of both online users. When I send a private message, the server finds the ip and a port and creates a socket from them.
The problem is that it doesn’t work correctly.
Here’s the code:
int portNumber = 4444;
String host = "192.168.0.100”;
Socket link;
try {
link = new Socket(host, portNumber);
// Then I set to already created PrintWriter the outputstream
out = new PrintWriter(link.getOutputStream(), true);
} catch (Exception e) {}
// Unfortunately the server freezes here (it doesn't show anything).
How to solve this problem? Where dod I make a mistake?
Thank you in advance.

You shouldn't create a new Socket to send a message. Instead, use a socket of an existing connection.
The sequence should be the following:
Client A connects to the server (server stores the connection as SocketA).
Client B connects to the server (server stores the connection as SocketB).
Server reads a private message from SocketA. The message is addressed to client B.
Server finds the existing socket for client B. It's SocketB.
Server sends the message into SocketB.

Java multiple multicast sockets in same group on same host and port

I'm using multicast to implement a simple discovery service. Several instances on the application must be able to run on the same host, I therefore end up with several multicast sockets that are members of the same group on the same host.
On Linux, this works as expected. Every instance of the application on a host receives the messages sent to the multicast address.
On Windows however, only the first application to join the multicast group receives the messages sent.
Here is the code I'm using to create my multicast socket:
socket = new MulticastSocket(PORT);
InetAddress group = InetAddress.getByName(ADDRESS);
socket.joinGroup(group);
socket.setTimeToLive(TTL);
Is there some option I need to set? I've tried scouring the API but I can't find anything.
EDIT: According to this website, I need to set SO_REUSEADDR, but this can only be set before the socket binds. It seems the constructor for MulticastSocket seems to bind the socket upon creation.

you do this by creating the socket with no arguments and then call bind on that object.
socket = new MulticastSocket();
socket.setReuseAddress(true);//redundant, already set with empty constructor
SocketAddress sockAddr = new InetSocketAddress(PORT);
socket.bind(sockAddr);
InetAddress group = InetAddress.getByName(ADDRESS);
socket.joinGroup(group);
socket.setTimeToLive(TTL);

How a threaded server continues to listen on the post for more incoming connections?

I am going through Sun's Java tutorial. I am in the lesson about sockets. There is the following code for a simple threaded server:
import java.net.*;
import java.io.*;
public class KKMultiServer {
public static void main(String[] args) throws IOException {
ServerSocket serverSocket = null;
boolean listening = true;
try {
serverSocket = new ServerSocket(4444);
} catch (IOException e) {
System.err.println("Could not listen on port: 4444.");
System.exit(-1);
}
while (listening)
new KKMultiServerThread(serverSocket.accept()).start();
serverSocket.close();
}
}
The server is said to "keep listening for more incoming connections". I just don't understand how it's possible; the line serverSocket.accept() constructs a new (client) Socket object which is, according to the tutorial "bound to the same local port and has its...". Well, how is it possible that the server is communicating with the client and listening to more incoming connections on the same port? As far as I know, if a port is used for some connection it is blocked and cannot be used for more things.
So what am I getting wrong here?

Well, a socket is not one-to-one based on a port, it is unique on a tuple of (address, port). A connection - the pair of the local and remote sockets involved in the communication - is used to demux incoming data from a port to the correct socket, allowing multiple sockets on one port. See Wikipedia. In other words, the relationship of sockets to ports are N-to-1

getting multiple connections on the same port is entirely possible as each TCP connection is a (local host, local port, remote host, remote port) tuple as long as at least 1 is different the connections are distinct and won't interfere (besides bandwidth drops)
clients attempting to connect to a server generally get a port assigned from the OS that is not used currently

Listening sockets work like a receptionist in on a business's phone switch. Everyone calls the switch number, and the receptionist responds to each incoming call on the switch line by having someone else handle the call on another line. Even though the receptionist can only take one call at a time, the switch line is tied up only very briefly because it is used only to establish a connection.
[...]TCP demultiplexes incoming segments using all four values that comprise the local and foreign addresses: destination IP address, destination port number, source IP address, and source port number. TCP cannot determine which process gets an incoming segment by looking at the destination port only. Also, the only one of the [various] endpoints at [a given port number] that will receive incoming connection requests is the one in the listen state. (p255, TCP-IP Illustrated Volume 1, W. Richard Stevens)
The last sentence in the above quote is the key to understanding.
Interestingly, a socket isn't really identified by the combination of IP address and port. This is unique only in context, where the context is either a particular connection or the listening state. Only one listener socket can bind to a particular IP/port combination.

The short and sweet answer is that the port is blocked for OTHER programs and processes. Only the program that opened the port can now listen on it. BUT it can listen to many different clients on the same port.
When a client connects, it creates a unique socket. A socket is comprised of the listening IP address and port (the one you opened) AND the calling IP address and port. Because the caller's IP address and port are always unique, each socket is unique and identifiable to your listener.
Even if I connected to your program twice from the same machine, my machine would select a new and random source port for each connection -- thus ensuring that we have a unique socket each time.

Based on this link
The accept method waits until a client starts up and requests a connection on the host and port of this server (in this example, the server is running on the hypothetical machine taranis on port 4444). When a connection is requested and successfully established, the accept method returns a new Socket object which is bound to the same local port and has its remote address and remote port set to that of the client.The server can communicate with the client over this new Socket and continue to listen for client connection requests on the original ServerSocket This particular version of the program doesn't listen for more client connection requests.
Here is SO discussion which may clear confusion about how single port handles multiple client calls Port and Socket SO discussion .
To put it in simple terms, most of the webservers listen on port 8080 and multiple clients will access same port to access your website.